Question: Let $g$ be a vector-valued function defined by $g(t)=(4\sin(\pi t),3t^3+3t^2+2t)$. Find $g$ 's second derivative $g''(t)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\left(4\pi\cos(\pi t),9t^2+6t+2\right)$ (Choice B) B $\left(-4\pi^2\sin(\pi t),18t+6\right)$ (Choice C) C $4\pi\cos(\pi t)-9t^2+6t+2$ (Choice D) D $\left(-4\sin(\pi t),18t+6\right)$
Answer: We are asked to find the second derivative of $g$. This means we need to differentiate $g$ twice. In other words, we differentiate $g$ once to find $g'$, and then differentiate $g'$ (which is a vector-valued function as well) to find $g''$. Recall that $g(t)=(4\sin(\pi t),3t^3+3t^2+2t)$. Therefore, $g'(t)=(4\pi\cos(\pi t),9t^2+6t+2)$. Now let's differentiate $g'(t)=(4\pi\cos(\pi t),9t^2+6t+2)$ to find $g''$. $g''(t)=(-4\pi^2\sin(\pi t),18t+6)$ In conclusion, $g''(t)=(-4\pi^2\sin(\pi t),18t+6)$.